What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x - 2} = \dfrac{6}{x - 2}$
Answer: Multiply both sides by $x - 2$ $ \dfrac{x^2 + 2}{x - 2} (x - 2) = \dfrac{6}{x - 2} (x - 2)$ $ x^2 + 2 = 6$ Subtract $6$ from both sides: $ x^2 + 2 - (6) = 6 - (6)$ $ x^2 + 2 - 6 = 0$ $ x^2 - 4 = 0$ Factor the expression: $ (x + 2)(x - 2) = 0$ Therefore $x = -2$ or $x = 2$ At $x = 2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 2$, it is an extraneous solution.